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%  Lecture 6
%  Source: handwritten notes (Mathpix-converted) + Kashlak STAT 571
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\section[Lecture 6 -- Integration and Convergence Theorems]{Lecture 6 \textemdash{} Integration and Convergence Theorems (MCT, Fatou, DCT)}
\label{sec:lec06}

Lectures~\ref{sec:lec04} (and~5) built up Lebesgue measure and the supply
of measurable functions; here we use that machinery to define the
\emph{Lebesgue integral} \(\int f\,d\mu\) for a measurable function
\(f:(\Omega,\Fcal,\mu)\to[-\infty,\infty]\), and state the three
convergence theorems on which essentially all of subsequent measure
theory rests: \emph{Monotone Convergence}, \emph{Fatou's Lemma}, and
\emph{Dominated Convergence}.

\medskip

\textbf{Notation.} For a sequence of functions \(f_i\) we write
\(f_i\uparrow f\) to mean \(f_i(\omega)\le f_{i+1}(\omega)\) for every
\(\omega\) and \(f_i(\omega)\to f(\omega)\) pointwise; analogously for
\(f_i\downarrow f\). We work in the extended real line
\([-\infty,\infty]\), with the conventions
\[
0\cdot\infty \;=\; 0, \qquad c\cdot\infty \;=\; \infty\ (c>0),
\qquad \infty-\infty \text{ undefined}.
\]
Every measurable \(f\) is split into its positive and negative parts
\[
f^+(\omega) \;=\; \max\{f(\omega),0\}, \qquad
f^-(\omega) \;=\; \max\{-f(\omega),0\},
\]
so that \(f=f^+-f^-\) and \(\lvert f\rvert=f^++f^-\).

\subsection{Building the integral via simple functions}

Recall from Lecture~5 that a \emph{simple function}
\(f=\sum_{i=1}^{p}x_i\indic_{B_i}\) (with \(B_i\in\Fcal\)) has the
unambiguous integral
\[
\int f\,d\mu \;=\; \sum_{i=1}^{p} x_i\,\mu(B_i),
\]
and that this integral is linear and monotone on non-negative simple
functions. The next theorem says that simple functions are dense, in the
strong sense that every measurable function is the increasing limit of
simple ones, so \emph{any} property linear in \(f\) and stable under
monotone limits which holds for indicators must hold for all measurable
functions.

\begin{theorem}{Approximation by simple functions on a $\pi$-system}{simple-density}
Let \((\Omega,\Fcal)\) be a measurable space and let \(\Acal\) be a
\(\pi\)-system that generates \(\Fcal\). Let \(\mathcal{V}\) be a linear space
of functions such that
\begin{enumerate}
  \item \(\indic_\Omega\in\mathcal{V}\) and \(\indic_A\in\mathcal{V}\) for every
        \(A\in\Acal\);
  \item whenever \(f_i\in\mathcal{V}\) and \(f_i\uparrow f\), the limit
        \(f\in\mathcal{V}\).
\end{enumerate}
Then \(\mathcal{V}\) contains every \(\Fcal\)-measurable function. Concretely,
for any non-negative measurable \(f\) the simple functions
\(f_i = 2^{-i}\lfloor 2^{i} f\rfloor\) satisfy \(f_i\uparrow f\).
\end{theorem}

\begin{definition}{Lebesgue integral of a measurable function}{lebesgue-integral}
Let \((\Omega,\Fcal,\mu)\) be a measure space and let \(f:\Omega\to[-\infty,\infty]\)
be measurable. For \(f\ge 0\) define
\[
\int f\,d\mu
   \;=\; \sup\!\left[\,\sum_{i}\Bigl\{\inf_{\omega\in A_i} f(\omega)\Bigr\}\mu(A_i)\,\right],
\]
where the supremum is taken over all finite measurable partitions
\(\{A_i\}\) of \(\Omega\); equivalently,
\(\int f\,d\mu = \sup\bigl\{\int g\,d\mu : g\text{ simple},\ 0\le g\le f\bigr\}\).
For general measurable \(f\), write \(f=f^+-f^-\) and set
\[
\int f\,d\mu \;=\; \int f^+\,d\mu \;-\; \int f^-\,d\mu,
\]
provided not both terms are infinite. We say \(f\) is \emph{integrable}
if both \(\int f^+\,d\mu\) and \(\int f^-\,d\mu\) are finite,
equivalently if \(\int\lvert f\rvert\,d\mu<\infty\).
\end{definition}

\begin{remark}
On a probability space \((\Omega,\Fcal,\P)\) the integral
\(\E X = \int X\,d\P\) is the \emph{expectation} of the random variable
\(X\); the integrability condition \(\int\lvert X\rvert\,d\P<\infty\) is
written \(X\in L^1(\Omega,\P)\).
\end{remark}

\begin{remark}
Following \cref{def:lebesgue-integral} the integral is linear and
monotone:
\[
\int(af+bg)\,d\mu \;=\; a\!\int f\,d\mu + b\!\int g\,d\mu,
\qquad f\le g \;\Longrightarrow\; \int f\,d\mu\le\int g\,d\mu,
\]
whenever the right-hand sides are defined. It also respects
\(\mu\)-null modifications, as the next theorem records.
\end{remark}

\begin{theorem}{Integrals ignore null sets}{integral-ae}
Let \((\Omega,\Fcal,\mu)\) be a measure space and let
\(f,g:\Omega\to[-\infty,\infty]\) be measurable with \(f=g\) almost
everywhere. Then \(f\) is integrable iff \(g\) is integrable, and in
that case \(\int f\,d\mu=\int g\,d\mu\).
\end{theorem}

\subsection{The three convergence theorems}

The substance of Lebesgue integration\,---\,its main advantage over the
Riemann integral\,---\,is the freedom with which we may exchange limits
and integrals. The three theorems below give three different sufficient
conditions; they form, in Koosis's words, ``the most important results
to learn'' in measure theory.

\begin{theorem}{Monotone Convergence (Beppo Levi)}{mct}
Let \((\Omega,\Fcal,\mu)\) be a measure space and let
\(\{f_i\}_{i=1}^{\infty}\) be measurable functions
\(\Omega\to\R\) with \(f_i\uparrow f\) almost everywhere and
\(\int f_1\,d\mu>-\infty\). Then \(f\) is measurable and
\[
\int f_i\,d\mu \;\uparrow\; \int f\,d\mu.
\]
\end{theorem}

\begin{remark}
It is enough that \(f_i\uparrow f\) hold \emph{almost everywhere}: by
\cref{thm:integral-ae} convergence may fail on a \(\mu\)-null set without
affecting the conclusion. A symmetric statement holds for
\(f_i\downarrow f\) provided \(\int f_1\,d\mu<\infty\).
\end{remark}

\begin{theorem}{Fatou's Lemma}{fatou}
Let \((\Omega,\Fcal,\mu)\) be a measure space and let
\(\{f_i\}_{i=1}^{\infty}\) be \emph{non-negative} measurable functions
\(\Omega\to\R\). Then
\[
\int \liminf_{i\to\infty} f_i\,d\mu
   \;\le\; \liminf_{i\to\infty}\int f_i\,d\mu.
\]
\end{theorem}

\begin{remark}
The inequality may be strict: take \(\Omega=\R\) with Lebesgue measure
and \(f_i = \indic_{[i,i+1]}\). Then \(f_i\to 0\) pointwise, so the
left-hand side is \(0\), while \(\int f_i\,d\mu=1\) for every \(i\) and
the right-hand side is~\(1\). Some hypothesis (monotonicity, domination,
\dots) is needed to upgrade ``\(\le\)'' to equality.
\end{remark}

\begin{theorem}{Dominated Convergence (Lebesgue)}{dct}
Let \((\Omega,\Fcal,\mu)\) be a measure space, let
\(\{f_i\}_{i=1}^{\infty}\) be measurable, and let \(g\) be absolutely
integrable (i.e.\ \(\int\lvert g\rvert\,d\mu<\infty\)). If
\begin{enumerate}
  \item \(\lvert f_i(\omega)\rvert\le g(\omega)\) for all \(i\) and all
        \(\omega\in\Omega\), and
  \item \(f_i(\omega)\to f(\omega)\) for each \(\omega\in\Omega\)
        (pointwise convergence),
\end{enumerate}
then \(f\) is absolutely integrable and
\[
\int f_i\,d\mu \;\longrightarrow\; \int f\,d\mu.
\]
\end{theorem}

\begin{remark}
On a probability space, \cref{thm:dct} is the standard tool for passing
limits inside expectations: if \(\lvert X_i\rvert\le Y\) with
\(\E Y<\infty\) and \(X_i\to X\) almost surely, then
\(\E X_i\to\E X\). Bounded convergence (\(g\equiv M\)) on a
\emph{finite} measure space is the simplest and most-used special case.
\end{remark}

\begin{corollary}{Bounded Convergence}{bdd-conv}
Let \(\mu(\Omega)<\infty\) and let \(\{f_i\}\) be measurable with
\(\lvert f_i\rvert\le M\) for some constant \(M<\infty\). If
\(f_i\to f\) pointwise (or a.e.), then \(\int f_i\,d\mu\to\int f\,d\mu\).
\end{corollary}